Log1/2(2x-3)>log1/2(2)
2x-3<2
2x<5
x<2,5
обл доп значений
2x-3>0
2x>3
x>1,5
ОТВЕТ
(1,5;2,5)
![(x-2)^4+(x^2-4)^2=0\\ (x-2)^4+(x-2)^2(x+2)^2=0\\ (x-2)^2((x-2)^2+(x+2)^2)=0\\ (x-2)^2(x^2-4x+4+x^2+4x+4)=0\\ (x-2)^2(2x^2+8)=0\\ (x-2)^2=0\\ x-2=0\\ x=2\\ 2x^2+8=0\\ x^2+4=0\\ x^2=-4\\ x\neq \sqrt{-4}](https://tex.z-dn.net/?f=%28x-2%29%5E4%2B%28x%5E2-4%29%5E2%3D0%5C%5C+%28x-2%29%5E4%2B%28x-2%29%5E2%28x%2B2%29%5E2%3D0%5C%5C+%28x-2%29%5E2%28%28x-2%29%5E2%2B%28x%2B2%29%5E2%29%3D0%5C%5C+%28x-2%29%5E2%28x%5E2-4x%2B4%2Bx%5E2%2B4x%2B4%29%3D0%5C%5C+%28x-2%29%5E2%282x%5E2%2B8%29%3D0%5C%5C+%28x-2%29%5E2%3D0%5C%5C+x-2%3D0%5C%5C+x%3D2%5C%5C+2x%5E2%2B8%3D0%5C%5C+x%5E2%2B4%3D0%5C%5C+x%5E2%3D-4%5C%5C+x%5Cneq+%5Csqrt%7B-4%7D)
Ответ: x=2.
![y=2x^4-x^2+4; \ x_0=-1\\\\ y-y_0=k(x-x_0)\\ y_0=2(-1)^4-(-1)^2+4=2-1+4=5\\ k=f'(x_0)=8x^3-2x=8(-1)^3-2(-1)=-6\\ y-5=-6(x-(-1))\\ y=-6x-1](https://tex.z-dn.net/?f=y%3D2x%5E4-x%5E2%2B4%3B+%5C+x_0%3D-1%5C%5C%5C%5C+y-y_0%3Dk%28x-x_0%29%5C%5C+y_0%3D2%28-1%29%5E4-%28-1%29%5E2%2B4%3D2-1%2B4%3D5%5C%5C+k%3Df%27%28x_0%29%3D8x%5E3-2x%3D8%28-1%29%5E3-2%28-1%29%3D-6%5C%5C+y-5%3D-6%28x-%28-1%29%29%5C%5C+y%3D-6x-1)
Уравнение касательной в точке с абсцисской -1 имеет вид y=-6x-1
Нок 13, 5,4 =260
13=13*1
5=5*1
4=2*2
нок=13*5*2*2=260
<span>3x³/(7-0,05y²)*3x³/(7+0,05y²)</span>